在等差数列an中a16+a17+a18=a9=-36,其前n项和为sn,求an的绝对值的前n项和Tn

Question

在等差数列an中a16+ a17+a18=a9=-36,其前n项和为sn,求an的绝对值的前n项和Tn

Answer 1

an =a1+(n-1)d
a16+a17+a18 = a9 = =-36
3a1+48d = -36 (1)
a1+8d = -36 (2)
(1)-3(2)
24d = 72
d = 3
a1= -60
an = -60+(n-1)3 = 3n - 63
Sn = (3n-123)n/2
an >0
3n-63 >0
n> 21
|an| = -an ; n≤21
= an ; n ≥22
for n≤21
|a1|+|a2|+...+|an|
= -(3n- 63 -60 ) n/2
= (123-3n)n/2
for n≥22
|a1|+|a2|+...+|an|
=S21 +a22+a23+.+an
= 630 + (3n-63)(n-20)/2